The answer is no. This might be kind of an esoteric question, so if anyone doesn't post by tomorrow I'll just post up my solution.

As promised:

Spoiler:

The answer is no.

Lemma: Every quasi-totally bounded (instead of finite, there is a countable net) metric space is separable.

Proof: For every let be the guaranteed countable -net and let . Clearly being the countable union of countable sets is countable. Also, we claim that it is dense in . To see this let and be given. By the Archimedean principle there exist some such that . So, then since cover it must be that for some . Thus, for that we have that from where the conclusion follows.

It follows since every compact metric space is totally bounded that every compact metric space is separable.

Also:

Lemma: Let be a space with dense. Then,

Proof: For a fixed define to be the set of all neighborhoods of and define . To see this is an injection we merely note that if then being guarantees there are neighboorhoods and such that and . In particular since every neighborhood of contains we have that and so and so . It follows that is an injection from where the conclusion follows.

Thus, either looking here, or noticing that and so and so by the above lemma cannot be turned into a separable space and thus not into a compact metric space.